A 19.6 g sample of ice at -10.0°C is mixed with 100.0 g of water at 75.4°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J g-1 °C-1, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.

Answer 1

qice=-qliq

q=delta H ice * n
q=S(liq)*Mass(liq)*(Tf-Ti)?

Help please!

I try setting the two equal to each other and plug in the numbers but I still get it wrong.

(6020 J/mol)*(19.6g)*(1mol/18g)=(4.18 J g-1 C-1)(100g)(Tf-75.4)